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Interestingly, there isn't for Gaussians (see my comment below). Other simple distributions tend to give biases towards the corners or axis. Perhaps the Gaussi

Interestingly, there isn't for Gaussians (see my comment below). Other simple distributions tend to give biases towards the corners or axis. Perhaps the Gaussian is unique in this regard? I'm not sure.

madcaptenor 11y ago
In three dimensions (to keep the notation reasonably simple) you need a distribution with density f such that f(x)f(y) f(z) is a function of x^2 + y^2 + z^2 (i. e. the distance from the origin). It looks like the Gaussian (up to scaling) is the only one.
OscarCunningham 11y ago
Yes, the fact that i.i.d. Gaussians give a rotationally symmetric joint distribution is a defining property of the Gaussian.
- darkmighty 11y ago
  Ah I suspected as much. Can you provide proof? I think I found one, but I'm not sure: Without loss of generality, take f(xi) = k1 * exp(-g(xi)) [1], for some g. Then we need the joint pdf to satisfy f(x1,...,xn) = k2 * exp(-h(R^2)), R=sum(xi^2)^1/2 (the R^2 and h(.) is w.l.g. too). So we get g(x1)+g(x2)=h(x1^2+x2^2). Then assuming the functions g and h analytic we end up needing g(x)= k * x^2, otherwise we get cross terms in the Taylor expansion that can't be cancelled out for all xi. Sounds good? [1] The function f trivially needs to be symmetric, justifying no loss of generality.
  - asdkashdkjashd 11y ago
    You have to assume that the one-dimensional marginals are Gaussian for otherwise the statement is not correct.
    - darkmighty 11y ago
      Hmm I'm not sure I get your point. I'm trying to prove that the joint pdf of N iid RV's is isotropic if and only if the RV's are gaussian. If I assume the pdfs are gaussian in the first place the proof isn't valid?
      - asdkashdkjashd 11y ago
        Oh, I see. I thought you were trying to prove something different.
  - madcaptenor 11y ago
    Perhaps coincidentally, someone just asked about this at math.stackexchange: http://math.stackexchange.com/questions/1255637/joint-pdf-of-n-1-i-i-d-random-variables-isotropic-if-and-only-if-they-are-cen http://math.stackexchange.com/questions/1255637/joint-pdf-of...
    - darkmighty 11y ago
      Yes, that was me :) I wanted to make sure I got a proof, since I didn't really find this elsewhere.

madcaptenor 11y ago

In three dimensions (to keep the notation reasonably simple) you need a distribution with density f such that f(x)f(y) f(z) is a function of x^2 + y^2 + z^2 (i. e. the distance from the origin). It looks like the Gaussian (up to scaling) is the only one.

OscarCunningham 11y ago

Yes, the fact that i.i.d. Gaussians give a rotationally symmetric joint distribution is a defining property of the Gaussian.

darkmighty 11y ago
Ah I suspected as much. Can you provide proof? I think I found one, but I'm not sure: Without loss of generality, take f(xi) = k1 * exp(-g(xi)) [1], for some g. Then we need the joint pdf to satisfy f(x1,...,xn) = k2 * exp(-h(R^2)), R=sum(xi^2)^1/2 (the R^2 and h(.) is w.l.g. too). So we get g(x1)+g(x2)=h(x1^2+x2^2). Then assuming the functions g and h analytic we end up needing g(x)= k * x^2, otherwise we get cross terms in the Taylor expansion that can't be cancelled out for all xi. Sounds good? [1] The function f trivially needs to be symmetric, justifying no loss of generality.
- asdkashdkjashd 11y ago
  You have to assume that the one-dimensional marginals are Gaussian for otherwise the statement is not correct.
  - darkmighty 11y ago
    Hmm I'm not sure I get your point. I'm trying to prove that the joint pdf of N iid RV's is isotropic if and only if the RV's are gaussian. If I assume the pdfs are gaussian in the first place the proof isn't valid?
    - asdkashdkjashd 11y ago
      Oh, I see. I thought you were trying to prove something different.
- madcaptenor 11y ago
  Perhaps coincidentally, someone just asked about this at math.stackexchange: http://math.stackexchange.com/questions/1255637/joint-pdf-of-n-1-i-i-d-random-variables-isotropic-if-and-only-if-they-are-cen http://math.stackexchange.com/questions/1255637/joint-pdf-of...
  - darkmighty 11y ago
    Yes, that was me :) I wanted to make sure I got a proof, since I didn't really find this elsewhere.